The freezing point of a 0.010 m aqueous solution of a nonvolatile solute is ?0.072°C. What would you expect the normal boiling point of this same solution to be?

have no clue, thanks for the thelp

Freezing point depression= N*kf* molality

you have freezing point depression, molality, and you know kf for water. Calculate N

Then:
bp = 100 + N*kb*m

yes, thanks al ot

erin erin answered
13 years ago

First you solve for I (vant hoff factor) in the freezing point eqn. You are provided molality, delta fp and can get the kf from a textbook (hint since its aqueous, use value for h2o).

Next, take this I value and plug it into the boiling point eqn with the given molality and the kb value from the text for h2o to solve for delta bp. Add 100 degrees (waters boiling point)

Step-by-Step Bot Step-by-Step Bot answered
8 months ago

To solve for the normal boiling point of the solution, we need to use the equation:

bp = 100 + N * kb * m

where
bp = boiling point
N = Van't Hoff factor (the number of particles into which the solute dissociates)
kb = boiling point elevation constant for the solvent
m = molality of the solution (moles of solute per kilogram of solvent).

Given that the freezing point depression is -0.072°C (which is equivalent to -0.072 K), we can calculate N using the formula:

Freezing point depression = N * kf * m

kf is the freezing point constant for the solvent, which is water in this case.

You mentioned that the solution is 0.010 m, so we substitute this value into the freezing point depression equation:

-0.072 K = N * kf * 0.010 m

We can rearrange the equation to solve for N:

N = (-0.072 K) / (kf * 0.010 m)

Now that we have the N value, we can substitute it into the boiling point equation to find the normal boiling point (bp):

bp = 100 + N * kb * m

The value of kb for water is 0.512 K kg/mol. Substituting the known values into the equation, we can find the normal boiling point.

Explain Bot Explain Bot answered
8 months ago

To find the normal boiling point of the solution, we can use the formula:

bp = 100 + N * kb * m

where bp is the boiling point in °C, N is the number of particles the solute dissociates into, kb is the boiling point elevation constant for the solvent (in this case, water), and m is the molality of the solution.

First, we need to calculate N, which represents the number of particles the solute dissociates into. Since it is stated that the solute is a nonvolatile solute, it does not evaporate or dissociate into ions in the solution. Therefore, N = 1.

Now, we need to find kb for water. The boiling point elevation constant, kb, for water is 0.512 °C/m.

Next, we need to calculate the molality of the solution. The molality is defined as the number of moles of solute per kilogram of solvent. In this case, we have a 0.010 m aqueous solution, which means that there are 0.010 moles of solute per kilogram of water.

Finally, we can substitute the values into the formula to find the normal boiling point:

bp = 100 + 1 * 0.512 * 0.010
= 100 + 0.00512
= 100.00512 °C

Therefore, we would expect the normal boiling point of this solution to be approximately 100.00512 °C.

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